In the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and electric field. For example, a uniform electric field
Esize 12{E} {} is produced by placing a potential difference (or voltage)
ΔVsize 12{V} {} across two parallel metal plates, labeled A and B. (See
[link] .) Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a more fundamental relationship between electric potential and electric field. From a physicist’s point of view, either
ΔVsize 12{V} {} or
Esize 12{E} {} can be used to describe any charge distribution.
ΔVsize 12{V} {} is most closely tied to energy, whereas
Esize 12{E} {} is most closely related to force.
ΔVsize 12{V} {} is a
scalar quantity and has no direction, while
Esize 12{E} {} is a
vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by
Esize 12{V} {} below.) The relationship between
ΔVsize 12{V} {} and
Esize 12{E} {} is revealed by calculating the work done by the force in moving a charge from point A to point B. But, as noted in
Electric Potential Energy: Potential Difference , this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an interesting special case.

The relationship between
Vsize 12{V} {} and
Esize 12{E} {} for parallel conducting plates is
E=V/dsize 12{E=V/d} {} . (Note that
ΔV=VABsize 12{ΔV=V rSub { size 8{"AB"} } } {} in magnitude. For a charge that is moved from plate A at higher potential to plate B at lower potential, a minus sign needs to be included as follows:
–ΔV=VA–VB=VAB . See the text for details.)

The work done by the electric field in
[link] to move a positive charge
qsize 12{q} {} from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

W=–ΔPE=–qΔV.size 12{W= - Δ"PE"= - qΔV} {}

The potential difference between points A and B is

–ΔV=–(VB–VA)=VA–VB=VAB.

Entering this into the expression for work yields

W=qVAB.size 12{W= ital "qV" rSub { size 8{ ital "AB"} } } {}

Work is
W=Fdcosθsize 12{W= ital "Fd""cos"?} {} ; here
cosθ=1 , since the path is parallel to the field, and so
W=Fd . Since
F=qE , we see that
W=qEd . Substituting this expression for work into the previous equation gives

qEd=qVAB.size 12{qEd= ital "qV" rSub { size 8{ ital "AB"} } } {}

The charge cancels, and so the voltage between points A and B is seen to be

VAB=EdE=VABd(uniformE- field only),

where
dsize 12{d} {} is the distance from A to B, or the distance between the plates in
[link] . Note that the above equation implies the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus the following relation among units is valid:

1 N/C=1 V/m.size 12{"1 N"/C="1 V"/m} {}

Voltage between points a and b

VAB=EdE=VABd(uniformE- field only),

where
dsize 12{d} {} is the distance from A to B, or the distance between the plates.

What is the highest voltage possible between two plates?

Dry air will support a maximum electric field strength of about
3.0×106V/msize 12{3×"10" rSup { size 8{6} } " V/m"} {} . Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?

Strategy

We are given the maximum electric field
Esize 12{E} {} between the plates and the distance
d between them. The equation
VAB=Edsize 12{V rSub { size 8{"AB"} } =Ed} {} can thus be used to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

VAB=Ed.size 12{V rSub { size 8{"AB"} } =Ed} {}

Entering the given values for
Esize 12{E} {} and
dsize 12{d} {} gives

VAB=(3.0×106V/m)(0.025 m)=7.5×104V

or

VAB=75 kV.size 12{V rSub { size 8{"AB"} } ="75" "kV"} {}

(The answer is quoted to only two digits, since the maximum field strength is approximate.)

Discussion

One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5 cm (1 in.) gap, or 150 kV for a 5 cm spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage will cause a spark if there are points on the surface, since points create greater fields than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up, say with static electricity, on dry days.

(in some cosmological theories) non-luminous material which is postulated to exist in space and which could take either of two forms: weakly interacting particles ( cold dark matter ) or high-energy randomly moving particles created soon after the Big Bang ( hot dark matter ).

Usman

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distance is the measurement of where you are and where you were
displacement is a measurement of the change in position

Shii

Thanks a lot

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